Perched upon Pinnacle Point, Peter picks up a peculiar piece of parchment (pictured below). "This might be a treasure map!" he exclaims as he flips it over, revealing a clue: $ $ P. Point. $673$ paces $ $ S. Cove. $861$ paces $ $ Under big rock “These must be the distances to the treasure!” Peter nearly wets himself. If Peter is currently facing towards Skeleton Cove, how many degrees to his left should he turn before walking $673$ paces to the treasure? Do not round during your calculations. Round your final answer to the nearest degree.
Solution: Converting the problem into geometrical terms Our problem can be modeled by the following triangle $\triangle ABC$, where we want to find $\angle A=\theta$. $A$ $\theta$ $C$ $B$ $989\text{ paces}$ $673\text{ paces}$ $861\text{ paces}$ Since we are given three side lengths, we can use the law of cosines. Using the law of cosines The law of cosines gives the following equation. $(BC)^2=(AB)^2+(AC)^2-2AB\!\cdot\! AC\!\cdot\!\cos(A)$ Solving the above equation for $\cos(A)$ gives the following equation. $\begin{aligned} \cos(A)&=\dfrac{(AB)^2+(AC)^2-(BC)^2}{2AB\!\cdot\! AC}\\\\ \cos(\theta)&=\dfrac{673^2+989^2-861^2}{2\cdot 673\cdot 989} \gray{\text{Substitute}}\\\\ \cos(\theta)&=\dfrac{689{,}729}{1{,}331{,}194}\\\\ \theta&=\cos^{-1}\left(\dfrac{689{,}729}{1{,}331{,}194}\right)\\\\ \theta&\approx 59^\circ \end{aligned}$ The answer Peter should turn $59^\circ$ to his left.